Need a *trivial* proof of an "obvious" combinatorial result I've just presented Enflo's theorem on the existence of a Banach space without an approximation property in my Functional Analysis class. The argument is not trivial by itself, but in order to emphasize the really interesting steps, I'd like to dismiss all the "obvious" lemmata in the most efficient way.
I have cleaned up the end quite a bit (if somebody is interested, the right choice of $k_m$ and $t_m$ is $t_{m+1}\in [t_1t_2\dots t_m,4t_1t_2\dots t_m]$ and $k_m=t_1t_2\dots t_{m-1}t_{m+1}$, which ensures that all interesting ratios are integers, so we can do a perfect fit without any dirty tail bounds and the corresponding estimates; also it is easier to use just random horizontal partitions instead of smart number-theoretic definitions). However some unpleasant pieces remain.
The one that irritates me most is the following:
Consider all $n-1$-element subsets $I$ of $\{1,2,\dots,2n\}$ and $m\in[1,2n-1]$. Let $N_o$ and $N_e$ be the numbers of $I$'s having odd and even number of elements in $\{1,\dots,m\}$ respectively. Then 
$$
|N_o-N_e|\le c_n(N_o+N_e)=c_n{2n\choose n-1}
$$
with some $c_n$ depending on $n$ only (so it should serve all $m$ simultaneously) such that $\sum_n \frac{c_n}n<+\infty$.
The sharp bound is, of course, $c_n=\frac 1n$ and Enflo gets it considering $m=1,2$ separately and using a trigonometric integral representation and Holder inequality for other $m$, on which I wasted about 35 minutes of the class time, but this should be just one-liner (OK, perhaps, three) presentable in under 10 minutes (preferably in under 5). Note that I don't care about sharp $c_n$ as long as the series above converges and will gladly trade its size for any simplification in the proof. 
So can we make this "obvious" fact formally obvious?
I'm asking here rather than on MO because it is an education question rather than a research one :-)
 A: Just an observation building upon Yly's answer which they suggested I put as an answer. Line 3 can be replaced with several more elementary lines.
First note that by line 2 we want to show
$2{{n-1}\choose{(m-1)/2}}/{{2n}\choose{m}} \leq 1/n $ for m odd.
$({{n-1}\choose{m/2}} - {{n-1}\choose{m/2 -1}})/{{2n}\choose{m}} \leq {{n}\choose{m/2}}/{{2n}\choose{m}}\leq 1/n$ for m even.
Certainly these are true for m=1,2, and by symmetry we just need to show them up to m=n. Now we show both expressions are decreasing in m up to n. First the odd case, replacing $m=2k+1$ we have
$2{{n-1}\choose{k+1}}/{{2n}\choose{2k+3}} = \frac{2k+3}{2n-2k-1}*2{{n-1}\choose{k}}/{{2n}\choose{2k+1}}$ 
While for the even case replacing $m=2k$ we have
${{n}\choose{k+1}}/{{2n}\choose{2k+2}} = \frac{2k+1}{2n-2k-1}*{{n}\choose{k}}/{{2n}\choose{2k}}$
A: I am probably too late, but here it is. Let $[x^k]f(x)$ denote a coefficient of $x^k$ in the polynomial $f(x)$.


*

*If we replace $m$ to $2n-m$, the difference $N_0-N_e$ is being multiplied by $(-1)^{n-1}$ by obvious reasons. Thus we may suppose that $1\leqslant m\leqslant n$.

*$N_o-N_e=[x^{n-1}](1-x)^m(1+x)^{2n-m}=[x^{n-1}](1-x^2)^{m}(1+x)^{2n-2m}$, thus $|N_o-N_e|\leqslant [x^{n-1}](1+x^2)^m(1+x)^{2n-2m}\leqslant 2^{2n-m}$, since any specific coefficient does not exceed the total sum of all coefficients. On the other hand, $N_0+N_e=\binom{2n}{n-1}\geqslant \binom{2n-1}{n-1}\geqslant \frac{2^{n-1}}{2n}$, since $\binom{2n-1}{n-1}$ is the maximal among binomial coefficients $\binom{2n-1}i,0\leqslant i\leqslant 2n-1$, and their sum equals $2^{2n-1}$. This implies that whenever $m\geqslant 10\log n$, we get $c_n<1/n$ for sure.

*Assume that $m< 10\log n$. We write $(1-x)^m(1+x)^{2n-m}=(1-x)(1-x)^{m-1}(1+x)^{2n-m}=\sum_{i=0}^{m-1}\binom{m-1}i(1-x)(-x)^i(1+x)^{2n-m}$. Thus $$|N_o-N_e|\leqslant \sum_{i=0}^{m-1}\binom{m-1}i\left|[x^{n-1}](1-x)x^i(1+x)^{2n-m}\right|.$$
At the same time, analogously we get
$$
N_o+N_e= \sum_{i=0}^{m-1}\binom{m-1}i[x^{n-1}](1+x)x^i(1+x)^{2n-m}.
$$
Therefore, if we prove the inequality $$\left|[x^{n-1}](1-x)x^i(1+x)^{2n-m}\right|\leqslant c_n [x^{n-1}](1+x)x^i(1+x)^{2n-m}$$
for certain constant $c_n$ and all $i\leqslant m<10\log n$, this constant $c_n$ works. This ratio of coefficients is, if I am not mistaken, $|2i+3-m|/(2n-m+1)=O(\log n/n)$, that is ok for you.
A: Update: This proof doesn't work as is, since $c_n$ should be independent of $m$. Maybe it can be massaged into finding a $c_n$ that does not depend on $m$, so I'll leave it here.

Well, one super-uninspired proof would be to write
$$
N_e-N_o=\sum_{k}{(-1)^k\binom{m}{k}\binom{2n-m}{n-1-k}},
$$
factor out $\dfrac{(2n-m)!}{(n-1)!(n+1)!}$ so as to clear denominators and common factors in the numerator,
and show that the $n^m$ terms in the remaining factor cancel out because
$$
\sum_{k}{(-1)^k\binom{m}{k}=0}.
$$
That means the remaining factor is a polynomial of degree at most $m-1$, so $c_n=O\left(\frac{1}{n}\right)$.
